ILL 1977 Problem 30

A triangle $ABC$ with $\angle A = 30^\circ$ and $\angle C = 54^\circ$ is given. On $BC$ a point $D$ is chosen such that $\angle CAD = 12^\circ.$ On $AB$ a point $E$ is chosen such that $\angle ACE = 6^\circ.$ Let $S$ be the point of intersection of $AD$ and $CE.$ Prove that $BS = BC.$